How to convert apparent power in kilovolt-amps (kVA) to real power in watts (W).
kVA to the watt calculation formula
The real power P in watts (W) is equal to 1000 times the apparent power S in kilovolt-amps (kVA), times the power factor PF:
P(W) = 1000 × S(kVA) × PF
So watts are equal to 1000 times kilovolt-amps times the power factor.
watts = 1000 × kilovolt-amps × PF
W = 1000 × kVA × PF
What is the real power in watts when the apparent power is 3 kVA and the power factor is 0.8?
P = 1000 × 3kVA × 0.8 = 2400W
ince watts is volts times amps, what is VA? VA (or volt-amps) is also volts times amps, the concept however has been expanded and extended cover some of the subleties of AC power.
For DC current:
VA = Watts (DC current).
In AC if the volts and amps are in phase (for example a resistive load) then the equation is also
VA=Watts (resistive load)
where V is the RMS voltage and A the RMS amperage.
But here is the difference. In AC, the volts and amps are not always in phase (meaning that the peak of the voltage curve is does not happen at the peak of the current curve). Think of pushing a swing. The greatest force is applied when the swing stops at the peak, but the greatest velocity is at the bottom of the arc. When you have reactive circuit elements the current can’t keep up with the voltage, but lags behind. So in AC, if the volts and amps are not precisely in phase you have to calculate the watts by multiplying the volts times the amps at each moment in time and take the average over time, i.e. the integral of V*A dt over a full cycle. The ratio between the VA (i.e. rms volts time rms amps) and Watts is called the power factor PF.
VA·PF = Watts (for any load, including inductive loads)
In other words, volt-amps x power factor = watts. Similarly, KVA*PF = KW,
Or kilovolt-amps times power factor equals kilowatts.
When you want to know how much the electricity is costing you, you use watts. When you are specifying equipment loads, fuses, and wiring sizes you use the VA, or the rms voltage and rms amperage. This is because VA considers the peak of both current and voltage, without taking into account if they happen at the same time or not.